Problem: $h(t) = 2t-5-4(f(t))$ $f(t) = -3t^{2}$ $ h(f(1)) = {?} $
Answer: First, let's solve for the value of the inner function, $f(1)$ . Then we'll know what to plug into the outer function. $f(1) = -3(1^{2})$ $f(1) = -3$ Now we know that $f(1) = -3$ . Let's solve for $h(f(1))$ , which is $h(-3)$ $h(-3) = (2)(-3)-5-4(f(-3))$ To solve for the value of $h$ , we need to solve for the value of $f(-3)$ $f(-3) = -3(-3)^{2}$ $f(-3) = -27$ That means $h(-3) = (2)(-3)-5+(-4)(-27)$ $h(-3) = 97$